Schrodinger time-independent equation in spherical polar co-ordinate system [ continue 2 ]

Cont...

to solve equation (15)

$$Put\; x= cos\theta$$

$$Now,\; \frac{d\Theta}{d\theta}=\frac{d\Theta}{dx}\cdot \frac{dx}{d\theta}=-sin\theta \frac{d\Theta}{dx}$$

Multiplying both sides by \(sin\theta\)

$$sin\theta\frac{d\Theta}{d\theta}=-sin^2\theta\frac{d\Theta}{dx}$$

Again,

$$\frac{d}{d\theta}\biggl[sin\theta \frac{d\Theta}{d\theta}\biggr]= \frac{d}{dx}\biggl[-sin^2\theta\frac{d\Theta}{dx}\biggr]\cdot \frac{dx}{d\theta}$$

$$=-sin\theta\frac{d}{dx}\biggl[-sin^2\theta \frac{d\Theta}{dx}\biggr]$$

$$=-sin\theta\frac{d}{dx}\biggl[-(1-cos^2\theta)\frac{d\Theta}{dx}\biggr]$$

$$or,\; \frac{d}{d\theta}\biggl[sin\theta\frac{d\Theta}{d\theta}\biggr]= sin\theta \frac{d}{dx}\biggl[(1-x^2)\frac{d\Theta}{dx}\biggr]\dotsm(17)$$

And

$$sin^2\theta=1-cos^2\theta=1-x^2\dotsm(18)$$

Using (17) and (18) in equation (15), we get,

$$\frac{d}{dx}\biggl[(1-x^2)\frac{d\Theta}{dx}\biggr]+\biggl[\beta-\frac{m^2}{1-x^2}\biggr]\Theta=0$$

$$or, \; (1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+\biggl[\beta-\frac{m^2}{1-x^2}\biggr]\Theta=0\dotsm(19)$$

For, m=0

$$(1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+\beta\Theta=0\dotsm(20)$$

For terminating series solution of above equation, The condition is that, value of \(\beta\) is found to be,

$$\beta=l(l+1)\dotsm(21)$$

With (21) equation (20 Is called Legendre's differential equation with l=0,1,2,3,.. Here 'l' is called orbital Quantum Number.

Substituting value of \(\beta\) in equation (19), we get,

$$(1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx}+\biggl[l(l+1)-\frac{m^2}{1-x^2}\biggr]\Theta=0\dotsm(22)$$

Equation (22) is called Associated Legendre'sdifferential equation solution of equation (22) gives associated Legendre's polynomial.

$$i.e.\; \Theta_L^m (x)= BP_l^m (x)\dotsm(23)$$

Where,

$$P_l^m(x)= (1-x^2)\frac m2 \frac{d^m}{dx^m}[P_l(x)]\Rightarrow Associated$$

$$P_l(x)= \frac{1}{2^l l!} \frac{d^l}{dx^l}\biggl[(x^2-1)^l\biggr]\Rightarrow legendre's\; polynomial$$

Equation (23) in terms of \(\theta\) can be written as,

$$\theta_l^m (\theta)=BP_l^m (cos\theta)\dotsm(24)$$

The condition of normalization is

$$\int_0^\pi [\Theta_l^m (\theta)]^*[ \Theta_l^m (\theta)] sin\theta d\theta=1$$

$$or,\; |B|^2 \int_0^\pi [ P_l^m (cos\theta)]^* [ P_l^m (cos\theta)]sin\theta d\theta=1$$

Using the orthogonality relation of Associated Legendre's polynomials.

$$|B|^2\biggl[\frac{2}{2l+1}\cdot \frac{(l+m)!}{(l-m)!}\biggr]=1$$

$$\therefore B= \biggl[\frac{2l+1}{2}\cdot \frac{(l-m)!}{(l+m)!}\biggr]^\frac12\dotsm(25)$$

Which is normalization constant for \(\Theta\)- wave function.

$$\Theta_l^m(\theta)=\biggl[\frac{2l+1}{2}\cdot \frac{(l-m)!}{(l+m)!}\biggr]^\frac12 P_l^m (cos\theta)\dotsm(26)$$

The total wave function for rigid rotator is given by,

$$\psi_l^m (\theta, \phi)= \biggl[\frac{2l+1}{4\pi}\cdot \frac{(l-m)!}{(l+m)!}\biggr]^\frac12 P_l^m (cos\theta)e^{im\phi}$$

$$m=0,\pm1,\pm2,...$$

$$l=0,1,2...$$

The solution is called spherical harmonics,

For l=0 , m=0

$$\psi_0^0(\theta, \phi)= \frac{1}{\sqrt{4\pi}}(independent\;of\;\theta\; and \;\phi$$

$$\rho= [\psi_0^0(\theta,\phi)]^*[\psi(\theta,\phi)]$$

$$=\frac{1}{4\pi}(S-orbital)$$

Reference:

1. Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
3. Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh.Quantum Mechanics.Chand & Company Ltd., 2002.